RBSE Solutions for Class 6 Maths Chapter 12 Algebra Ex 12.2 is part of RBSE Solutions for Class 6 Maths. Here we have given Rajasthan Board RBSE Class 6 Maths Chapter 12 Algebra Exercise 12.2.
| Board | RBSE |
| Textbook | SIERT, Rajasthan |
| Class | Class 6 |
| Subject | Maths |
| Chapter | Chapter 12 |
| Chapter Name | Algebra |
| Exercise | Ex 12.2 |
| Number of Questions | 9 |
| Category | RBSE Solutions |
Rajasthan Board RBSE Class 6 Maths Chapter 12 Algebra Ex 12.2
Question 1.
With the numbers 3, 7 and 4, form (RBSESolutions.com) arithmetic expressions using
(i) Only addition and subtraction operations.
(ii) Only multiplication and addition operations.
Solution:
(i) 3 + 7 + 4,
3 – 7 + 4,
3 + 7 – 4,
– 3 + 7 + 4,
– 3 + 7 – 4,
– 3 – 7 + 4,
– 3 – 7 – 4.
(ii) 3 × 7 + 4,
3 × 7 × 4,
3 + 7 × 4,
3 + 4 × 7,
3 × 4 + 7.
Question 2.
For each of the below expressions, mention (RBSESolutions.com) whether it is an arithmetic expresion or algebric expression?
(i) (3x + 5)
(ii) 5 × 4 + 7
(iii) 3 + 4 × 3 + 5
(iv) 2x + 1
(v) \(\frac { x }{ 2 } \) + 5 – x
(vi) 3x
Solution:
Algebric expression :
(i) 3x + 5
(iv) 2x + 1
(v) \(\frac { x }{ 2 } \) + 5 – x
(vi) 3x
Arithmetic expression :
(ii) 5 × 4 +7
(ii) 3 + 4 × 3 + 5
Question 3.
Observe the expressions in the table carefully. Mention, (RBSESolutions.com) what all operation are used to form that expression by putting the sign of right/wrong in the table.
Solution:
Question 4.
Write algebraic expression (RBSESolutions.com) for the following situations :
(i) 7 added to a
(ii) 10 subtracted from b
(iii) x multiplied by 4
(iv) x divided by 4
(v) x subtracted from 7
(vi) 10 divided by q.
Solution:
Question 5.
Give expressions for (RBSESolutions.com) the following cases :
(i) 15 added to 2n
(ii) 15 subtracted from 2x
(iii) 3 added to twice of p
(iv) 3 subtracted from twice of q
(v) 11 subtracted from the product of y and 5
(vi) 11 added to the product of z and – 3
Solution:
Question 6.
Form algebraic expressions (RBSESolutions.com) using q, 5 and – 3.
Solution:
(i) q + 5 – 3
(ii) q – 5 – 3
(iii) -q + 5 – 3
(iv) q ÷ 5 + (- 3)
(v) 5q – 3
(vi) – 5q – 3
(vii) \(\frac { 5 }{ q } -3\)
(viii) \(\frac { -q }{ 5 } -3\)
(ix) 5 – 3q
(x) – 5 – 3q
Question 7.
Nathu has Rs. x with (RBSESolutions.com) him Then,
(i) How much money do Bina have if she owns twice as much as Nathu does?
(ii) How much money is Nathu left with after buying books worth Rs. 150?
(iii) How much money do Seema have if she owns half as much as Nathu has initially?
(iv) How much money do Milli have if she owns thrice as much as Nathu does?
Solution:
(i) Nathu has Rs. = x
Bina has money = Twice money as much
as Nathu’s money = 2 × x = Rs. 2x
(ii) Nathu purchased books of = Rs. 150
money remains = Rs. (x – 150)
(iii) Nathu has intial money = Rs. x
Seema has money = half money as much
as Nathu’s initial money = \(\frac { x }{ 2 }\) Rs.
(iv) Milli has money = Thrice money as much
as Nath’s money = 3 × x = Rs. 3x
Question 8.
The height of a triangle is 5 more than (RBSESolutions.com) twice of its base. What is its height if base is b?
Solution:
Base of triangle= b unit
Height of triangle = 5 more than twice of base
= 2 × b + 5 = (2b + 5) unit
Question 9.
The present age of Vimal is p years. then
(i) How old was he 10 years ago?
(ii) How old will Vimal be 5 years from now?
(iii) Vimal’s aunt is (RBSESolutions.com) thrice as old as vimal. How old is vimal’s aunt?
(iv) Age of Vimal’s mother is 5 years less than twice of vimal’s age. How old is his mother?
Solution:
Vimal’s present age = P years
(i) His age 10 years ago = (p – 10) years
(ii) Vimal’s age after 5 years from now = (p + 5) years
(iii) Vimal’s aunt age = 3 × p = 3p years
(iv) Vimal’s mother age = 5 years less than twice of vimal’s age = 2 × p – 5 = (2p – 5) years
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